Termination w.r.t. Q proof of TRS/higher-order/AProVE_HOTypeEx3.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

app(app(f, 0), n) → app(app(hd, app(app(map, f), app(app(cons, 0), nil))), n)
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

app(app(f, 0), n) → app(app(hd, app(app(map, f), app(app(cons, 0), nil))), n)
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(f, 0), n) → APP(map, f)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(f, 0), n) → APP(app(cons, 0), nil)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(f, 0), n) → APP(app(map, f), app(app(cons, 0), nil))
APP(app(f, 0), n) → APP(app(hd, app(app(map, f), app(app(cons, 0), nil))), n)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(f, 0), n) → APP(cons, 0)
APP(app(f, 0), n) → APP(hd, app(app(map, f), app(app(cons, 0), nil)))

The TRS R consists of the following rules:

app(app(f, 0), n) → app(app(hd, app(app(map, f), app(app(cons, 0), nil))), n)
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

Q DP problem:
The TRS P consists of the following rules:

APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(f, 0), n) → APP(map, f)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(f, 0), n) → APP(app(cons, 0), nil)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(f, 0), n) → APP(app(map, f), app(app(cons, 0), nil))
APP(app(f, 0), n) → APP(app(hd, app(app(map, f), app(app(cons, 0), nil))), n)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(f, 0), n) → APP(cons, 0)
APP(app(f, 0), n) → APP(hd, app(app(map, f), app(app(cons, 0), nil)))

The TRS R consists of the following rules:

app(app(f, 0), n) → app(app(hd, app(app(map, f), app(app(cons, 0), nil))), n)
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(f, 0), n) → APP(map, f)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(f, 0), n) → APP(app(cons, 0), nil)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(f, 0), n) → APP(app(hd, app(app(map, f), app(app(cons, 0), nil))), n)
APP(app(f, 0), n) → APP(app(map, f), app(app(cons, 0), nil))
APP(app(f, 0), n) → APP(hd, app(app(map, f), app(app(cons, 0), nil)))
APP(app(f, 0), n) → APP(cons, 0)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(f, 0), n) → app(app(hd, app(app(map, f), app(app(cons, 0), nil))), n)
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(f, 0), n) → APP(app(cons, 0), nil)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(f, 0), n) → APP(app(hd, app(app(map, f), app(app(cons, 0), nil))), n)
APP(app(f, 0), n) → APP(app(map, f), app(app(cons, 0), nil))

The TRS R consists of the following rules:

app(app(f, 0), n) → app(app(hd, app(app(map, f), app(app(cons, 0), nil))), n)
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.